\( 20 \, \text{N} \)
\( 10 \, \text{N} \)
Provided data:
Normal force equals object weight: \[ N = m \cdot g = 5 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 49 \, \text{N} \]
Friction force formula: \[ f_{\text{friction}} = \mu \cdot N = 0.4 \times 49 \, \text{N} = 19.6 \, \text{N} \] Rounded to the nearest integer: \[ f_{\text{friction}} = 20 \, \text{N} \]
The block experiences a friction force of \( 20 \, \text{N} \).
Option 1: 20 N

A block of mass m is placed on a surface having vertical cross section given by \(y=\frac{x^2}{4}\). If coefficient of friction is 0.5, the maximum height above the ground at which block can be placed without slipping is: